Using Nernst equation: Ecell = E°cell - (0.059/n)log(Q). Increasing [Zn²⁺] increases Q, making the log term positive, which decreases Ecell. ΔE = -(0.059/2)log(10) = -0.0295 ≈ -0.0296 V.

At cathode: Na⁺ + e⁻ → Na; moles of Na = 2.3/23 = 0.1 mol. At anode: 2Cl⁻ → Cl₂ + 2e⁻; for 0.1 mol Na, electrons = 0.1 mol, so Cl₂ moles = 0.1/2 = 0.05 mol. Volume at STP = 0.05 × 22.4 = 1.12 L.

The species with highest reduction potential (+1.36 V) is Cl₂, making it the strongest oxidizing agent. Higher E° values indicate greater tendency to accept electrons.

As the reaction proceeds, product concentrations increase while reactant concentrations decrease, reducing the driving force according to Nernst equation: E = E° - (0.059/n)log(Q).

At cathode: 2H⁺ + 2e⁻ → H₂ (1 mol H₂). At anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ (0.5 mol O₂). With 2 mol e⁻: 1 mol H₂ (11.2 L) and 0.5 mol O₂ (5.6 L).

In electrorefining, impure copper acts as anode and undergoes oxidation. Pure copper deposits at cathode. More reactive impurities go into solution.

The standard free energy change ΔG° = -nFE°cell, where n is moles of electrons, F is Faraday constant, and E° is standard cell potential.

At cathode with copper electrodes in CuSO₄: Cu²⁺ ions are preferentially reduced as their reduction potential (+0.34 V) is higher than H⁺ (-0.83 V).

Λm = κ × 1000/c (molar conductance), Λ = κ × 1000/(c/n) (equivalent conductance). Therefore, Λm = Λ/n or Λ = Λm × n.

Charge = I × t = 5 × 1930 = 9650 C. Moles of e⁻ = 9650/96500 = 0.1 mol.